Question

The number of terms between 105 and 1000 which is divisible by 7 is
A. 142
B. 128
C. 127
D. None of these

Answer 1

Hint: First of all, find out the first number and the last number that is divisible by 7. Since we have to find all the numbers that are divisible by 7 between 105 and 1000, the numbers will form an arithmetic progression (A.P.) with common difference 7. Use the last term of A.P. formed to find the total number of terms.

Complete step-by-step answer:
We will find the first number that will be divisible by 7 after 105.
As, we have to calculate the number of terms between 105 and 1000, we will not include 105 and 1000.
Therefore, the first number divisible by 7 after 105 is 112. Now, we will find the last term for the sequence.
994 is the last term that is divisible by 7 before 1000.
Hence, the A.P. is formed, where first term is 112 , last term is 994 and common difference is 7.
Use the formula of ${n^{th}}$ term, ${a_n} = a + \left( {n - 1} \right)d$ to find the number of terms in A.P.
Substitute 112 for $a$, 994 for ${a_n}$ and 7 for $d$ in the formula for ${n^{th}}$ term, $ \Rightarrow{a_n} = a + \left( {n - 1} \right)d$ to determine the value of $d$ .
$ \Rightarrow 994 = 112 + \left( {n - 1} \right)7$
$ \Rightarrow 994 - 112 = \left( {n - 1} \right)7$
$ \Rightarrow882 = \left( {n - 1} \right)7$
$ \Rightarrow
  \dfrac{{882}}{7} = \left( {n - 1} \right) \\
\Rightarrow \left( {n - 1} \right) = 126 \\
 \Rightarrow n = 126 + 1 \\

 \Rightarrow n = 127 \\
$
Hence, there are 127 terms between 105 and 1000 that are divisible by 7.
Hence, option C is correct.

Note: Do not include terms 105 and 1000 as we have to calculate the number of terms between 105 and 1000. The first number and the last number is calculated by hit and trial. The numbers divisible by 7 will form an A.P. because the common difference of the terms will be 7.